Re: Limits

I have another problem regarding this.. This comes as the 2nd part of the above problem.

Let f be differentiable on (0,infinity) using the mean value theorem and the result proven in the above question prove that of f and f' both are strictly increasing on (0,infinity) then limit of f(x) when x goes to infinity equals infinity.

Well as f is differentiable on (0,infinity) we can say f is continuous on (0,infinity).. therefore we can apply the mean value theorem for this.

Re: Limits

Well as f is differentiable on (0,infinity) we can say f is continuous on (0,infinity).. therefore we can apply the mean value theorem for this.

You are correct that since is differentiable on , it is continuous on the same interval. However, what is required to apply the Mean Value Theorem (I will abbreviate MVT from now on)?

The statement of the MVT requires that be continuous on a closed and bounded interval and differentiable on the interior of the same interval (the same interval without its endpoints is an open interval). So, you cannot apply the MVT to (0,M). Instead, you need a closed interval. Well, let's create a closed interval. Choose arbitrary points . We know is continuous on and differentiable on , so we can apply the MVT on this interval.

Originally Posted by Kristen111111111111111111

there exist some 'c' in (0,M) s.t f'(c) = [f(M)-f(0)]/M > 0

So, after you apply the MVT on this interval, you get some as you stated (only I am using different interval to find it and a different variable). To state a little more formally, by the MVT, there exists such that . Since we chose and with and we are given that is strictly increasing, this implies that both the numerator and denominator are positive. Hence, their quotient is positive. So, we now have a point with . Let and . You have , so if you can show that for all , then you can apply the first result to prove this 2nd part of your problem.